This chapter can be considered to be the heart of the course. We now use all the machinery that has been developed to this point in order to study the most impor-tant class of functions in analysis, namely, continuous functions. In Section 5.3, the fundamental properties of continuous functions are proved, and this section is the most important of this chapter. Sufficient time should be spent on it to allow adequate study of the proofs and examples. Section 5.4 on uniform continuity is also an important section. Section 5.5 contains a different approach to the basic theorems in Sections 5.3 and 5.4, using the idea of a “gauge”.
The results on monotone functions in Section 5.6 are interesting, but they are not central to this course and these results will not be used often in later parts of this book.
Section 5.1
This important section is absolutely basic to everything that will follow. Every effort should be made to have the students master the notions presented here.
They should memorize the definition of continuity and its various equivalents, and they should study the examples very carefully.
Sample Assignment: Exercises 1, 3, 4(a,b), 5, 7, 11, 12, 13.
Partial Solutions:
3. We will establish the continuity of h at b. Since f is continuous at b, given ε > 0 there exists δ1> 0 such that if b− δ1< x < b, then|f(x) − f(b)| < ε. Sim-ilarly, there exists δ2> 0 such that if b < x < b + δ2, then|g(x) − g(b)| < ε. Let δ := inf{δ1, δ2} so that |h(x) − h(b)| < ε for |x − b| < δ, whence h is continuous at b.
4. (a) Continuous if x = 0, ±1, ±2, . . . , (b) Continuous if x = ±1, ±2, . . . , (c) Continuous if sin x = 0, 1,
(d) Continuous if x = 0, ±1, ±1/2, . . . . 5. Yes. Define f (2) := lim
x→2f (x) = 5.
6. Given ε > 0, choose δ > 0 such that if x∈ Vδ(c)∩ A, then |f(x) − f(c)| < ε/2.
Then if y∈ Vδ(c)∩ A, we have |f(y) − f(x)| ≤ |f(x) − f(c)| + |f(c) − f(y)| <
ε/2 + ε/2 = ε.
7. Let ε := f (c)/2, and let δ > 0 be such that if|x − c| < δ, then |f(x) − f(c)| < ε, which implies that f (x) > f (c)− ε = f(c)/2 > 0.
8. Since f is continuous at x, we have f (x) = lim(f (xn)) = 0. Thus x∈ S.
33
34 Bartle and Sherbert
9. (a) If|f(x) − f(c)| < ε for x ∈ Vδ∩ B, then |g(x) − g(c)| = |f(x) − f(c)| < ε for x∈ Vδ(c)∩ A.
(b) Let f = sgn on B := [0, 1], g = sgn on A := (0, 1] and c = 0.
10. Note that |x| − |c| ≤|x− c|.
11. Let c∈ R be given and let ε > 0. If |x − c| < ε/K, then |f(x) − f(c)| ≤ K|x − c| < K(ε/K) = ε.
12. If x is irrational, then by the Density Theorem 2.4.8 there exists a sequence (rn) of rational numbers that converges to x. Then f (x) = lim(f (rn)) = 0.
13. Since |g(x) − 6| ≤ sup{|2x − 6|, |x − 3|} = 2|x − 3|, then g is continuous at x = 3. If c = 3, let (xn) be a sequence of rational numbers converging to c and let (yn) be a sequence of irrational numbers converging to c. Then lim(g(xn)) = 2c = c + 3 = lim(g(yn)), so g is not continuous at c.
14. Let c∈ A. If k is continuous at c, it follows from 4.2.2 that k is bounded on some neighborhood (c− δ, c + δ). Let m ∈ N be given; then there exists a prime number p such that 1/p < δ and p≥ m. (Why?) There must be at least one rational number q/p with c− δ < q/p < c + δ; otherwise there exists an integer q0 such that q0/p≤ c − δ and c + δ ≤ (q0+ 1)/p, which implies that 2δ≤ 1/p, a contradiction. We conclude that k(x) = p ≥ m for at least one point x∈ (c − δ, c + δ). But this is a contradiction.
15. Let In:= (0, 1/n] for n∈ N. Show that (sup f(In)) is a decreasing sequence and (inf f (In)) is an increasing sequence. If lim(sup f (In)) = lim(inf f (In)), then lim
x→0f exists. Let xn, yn∈ In be such that f (xn) > sup f (In)− 1/n and f (yn) < inf f (In) + 1/n.
Section 5.2
Note the similarity of this section with Sections 4.2 and 3.2. However, Theorem 5.2.6 concerning composite functions is a new result, and an important one. Its importance may be suggested by the fact, noted in 5.2.8, that it implies several of the earlier results.
The significance of this section should be clear: it enables us to establish the continuity of many functions.
Sample Assignment: Exercises 1, 3, 5, 6, 10, 12, 13.
Partial Solutions:
1. (a) Continuous on R, (b) Continuous for x≥ 0, (c) Continuous for x = 0, (d) Continuous on R.
2. Use 5.2.1(a) and Induction; or, use 5.2.8 with g(x) := xn.
3. Let f be the Dirichlet discontinuous function (Example 5.1.6(g)) and let g(x) := 1− f(x).
4. Continuous at every noninteger.
5. The function g is not continuous at 1 = f (0).
6. Given ε > 0, there exists δ1> 0 such that if|y − b| < δ1, then |g(y) − g(b)| < ε.
Further, there exists δ > 0 such that if 0 <|x − c| < δ, then |f(x) − b| < δ1. Hence, if 0 < |x − c| < δ, then we have |(g ◦ f)(x) − g(b)| < ε, so that
x→clim(g◦ f)(x) = g(b).
7. Let f (x) := 1 if x is rational, and f (x) :=−1 if x is irrational.
8. Yes. Given x∈ R, let (rn) be a sequence of rational numbers with rn→ x.
9. Show that an arbitrary real number is the limit of a sequence of numbers of the form m/2n, where m∈ Z, n ∈ N.
10. If c∈ P , then f(c) > 0. Now apply Theorem 4.2.9.
11. If h(x) := f (x)− g(x), then h is continuous and S = {x ∈ R : h(x) 0}.
12. First show that f (0) = 0 and f (−x) = −f(x) for all x ∈ R; then note that f (x− x0) = f (x)− f(x0). Consequently f is continuous at the point x0 if and only if it is continuous at 0. Thus, if f is continuous at x0, then it is continuous at 0, and hence everywhere.
13. First show that f (0) = 0 and (by Induction) that f (x) = cx for x∈ N, and hence also for x∈ Z. Next show that f(x) = cx for x ∈ Q. Finally, if x /∈ Q, let x = lim(rn) for some sequence in Q.
14. First show that either g(0) = 0 or g(0) = 1. Next, if g(α) = 0 for some α∈ R and if x ∈ R, let y := x − α so that x = α + y; hence g(x) = g(α + y) = g(α)g(y) = 0. Thus, if g(α) = 0 for some α, then it follows that g(x) = 0 for all x∈ R.
Now suppose that g(0) = 1 so that g(c) = 0 for any c ∈ R. If g is continuous at 0, then given ε > 0 there exists δ > 0 such that if|h| < δ, then |g(h) − 1| <
ε/|g(c)|. Since g(c+h)−g(c) = g(c)(g(h)−1), it follows that |g(c+h)−g(c)| =
|g(c)g(h) − 1| < ε, provided |h| < δ. Therefore g is continuous at c.
15. If f (x)≥ g(x), then both expressions given h(x) = f(x); and if f(x) ≤ g(x), then h(x) = g(x) in both cases.
Section 5.3
In this section, we establish some very important properties of continuous func-tions. Unfortunately, students often regard these properties as being “obvious”, so that one must convince them that if the hypotheses of the theorems are dropped, then the conclusions may not hold. Thus, for example, if any one of the three hypotheses [(i) I is closed, (ii) I is bounded, (iii) f is continuous at every point of I] of Theorem 5.3.2 is dropped, then the conclusion that f is bounded may not hold, even though the other two hypotheses are retained. Similarly for Theorems 5.3.4 and 5.3.9. Thus, each theorem must be accompanied by examples.
In 5.3.7, we do not assume that I is a closed bounded interval, but we work within a closed bounded subinterval of I.
36 Bartle and Sherbert
The proofs of Theorems 5.3.2 and 5.3.4 presented here are based on the Bolzano-Weierstrass Theorem. In Section 5.5 different proofs are presented based on the concept of a “gauge”. In Chapter 11 these theorems are extended to general
“compact” sets in R by using the Heine-Borel Theorem.
Students often misunderstand Theorems 5.3.9 and 5.3.10, believing that the image of an interval with endpoints f (a), f (b). Consequently, Figure 5.3.3 should be stressed in an attempt to dispell this misconception. Also, examples can be given to show that the continuous image of an interval (a, b) can be any type of interval, and not necessarily an open interval or a bounded interval.
Sample Assignment: Exercises 1, 3, 5, 6, 7, 8, 10, 13, 15.
Partial Solutions:
1. Apply either the Boundedness Theorem 5.3.2 to 1/f , or the Maximum-Minimum Theorem 5.3.4 to conclude that inf f (I) > 0.
Alternatively, if xn∈ I such that 0 < f(xn) < 1/n, then there is a subse-quence (xnk) that converges to a point x0∈ I. Since f(x0) = lim(f (xnk)) = 0, we have a contradiction.
2. If f (xn) = g(xn) and lim(xn) = x0, then f (x0) = lim(f (xn)) = lim(g(xn)) = g(x0).
3. Let x1 be arbitrary and let x2∈ I be such that |f(x2)| ≤12|f(x1)|. By Induc-tion, choose xn+1 such that |f(xn+1)| ≤12|f(xn)| ≤ 1
2
n
|f(x1)|. Apply the Bolzano-Weierstrass Theorem to obtain a subsequence that converges to some c∈ I. Now show that f(c) = 0.
Alternatively, show that if the minimum value of |f| on I is not 0, then a contradiction arises.
4. Suppose that p has odd degree n and that the coefficient anof xnis positive.
By 4.3.16, we have lim
x→∞p(x) =∞ and lim
x→−∞p(x) =−∞. Hence p(α) < 0 for some α < 0 and p(β) > 0 for some β > 0. Therefore there is a zero of p in [α, β].
5. In the intervals [1.035, 1.040] and [−7.026, −7.025].
6. Note that g(0) = f (0)− f(1/2) and g(1/2) = f(1/2) − f(1) = −g(0). Hence there is a zero of g at some c∈ [0, 1/2]. But if 0 = g(c) = f(c) − f(c + 1/2), then we have f (c) = f (c + 1/2).
7. In the interval [0.7390, 0.7391].
8. In the interval [1.4687, 1.4765].
9. (a) 1, (b) 6.
10. 1/2n< 10−5 implies that n > (5 ln 10)/ ln 2≈ 16.61. Take n = 17.
11. If f (w) < 0, then it follows from Theorem 4.2.9 that there exists a δ-neighborhood Vδ(w) such that f (x) < 0 for all x∈ Vδ(w). But since w < b,
this contradicts the fact that w = sup W . There is a similar contradiction if we assume that f (w) > 0. Therefore f (w) = 0.
12. Since f (π/4) < 1 while f (0) = 1 and f (π/2) > 1, it follows that x0 ∈ (0, π/2).
If cos x0 > x20, then there exists a δ-neighborhood Vδ(x0) ⊆ I on which f (x) = cos x, so that x0 is not an absolute minimum point for f .
13. If f (x) = 0 for all x∈ R, then all is trivial; hence, assume that f takes on some nonzero values. To be specific, suppose f (c) > 0 and let ε :=12f (c), and let M > 0 be such that |f(x)| < ε provided |x| > M. By Theorem 5.3.4, there exists c∗∈ [−M, M] such that f(c∗)≥ f(x) for all x ∈ [−M, M] and we deduce that f (c∗)≥ f(x) for all x ∈ R. To see that a minimum value need not be attained, consider f (x) := 1/(x2+ 1).
14. Apply Theorem 4.2.9 to β− f(x).
15. If 0 < a < b≤ ∞, then f((a, b)) = (a2, b2); if −∞ ≤ a < b < 0, then f((a, b)) = (b2, a2). If a < 0 < b, then f ((a, b)) is not an open interval, but equals [0, c) where c := sup{a2, b2}. Images of closed intervals are treated similarly.
16. For example, if a < 0 < b and c := inf{1/(a2+ 1), 1/(b2+ 1)}, then g((a, b)) = (c, 1]. If 0 < a < b, then g((a, b)) = (1/(b2+ 1), 1/(a2+ 1)). Also g([−1, 1]) = [1/2, 1]. If a < b, then h((a, b)) = (a3, b3) and h((a, b]) = (a3, b3].
17. Yes. Use the Density Theorem 2.4.8.
18. If f is not bounded on I, then for each n∈ N there exists xn∈ I such that
|f(xn)| ≥ n. Then a subsequence of (xn) converges to x0∈ I. The assumption that f is bounded on a neighborhood of x0 leads to a contradiction.
19. Consider g(x) := 1/x for x∈ J := (0, 1).
Section 5.4
The idea of uniform continuity is a subtle one that often causes difficulties for stu-dents. The point, of course, is that for a uniformly continuous function f : A→ R, the δ can be chosen to depend only on ε and not on the points in A. The Uniform Continuity Theorem 5.4.3 guarantees that every continuous function on a closed bounded interval is uniformly continuous; however, a continuous func-tion defined on an interval may be uniformly continuous even when the interval is not closed and bounded. For example, every Lipschitz function is uniformly continuous, no matter what the nature of its domain is. A condition for a function to be uniformly continuous on a bounded open interval is given in 5.4.8.
The extension of the Uniform Continuity Theorem to compact sets in given in Chapter 11.
One interesting application of uniform continuity is the approximation of continuous functions by “simpler” functions. Consequently we have included a brief discussion of this topic here. The Weierstrass Approximation Theorem 5.4.14 is a fundamental result in this area and we have stated it without proof.
38 Bartle and Sherbert
Sample Assignment: Exercises 1, 2, 3, 6, 7, 8, 11, 12, 15.
Partial Solutions:
1. Since 1/x− 1/u = (u − x)/xu, it follows that |1/x − 1/u| ≤ (1/a2)|x − u| for x, u∈ [a, ∞).
2. If x, u ≥ 1, then |1/x2− 1/u2| = (1/x2u + 1/xu2)|x − u| ≤ 2|x − u|, and it follows that f is uniformly continuous on [1,∞). If xn:= 1/n, un:= 1/(n + 1), then |xn− un| → 0 but |f(xn)− f(un)| = 2n + 1 ≥ 1 for all n, so f is not uni-formly continuous on (0,∞).
3. (a) Let xn:= n + 1/n, un:= n. Then |xn− un| → 0, but f(xn)− f(un) = 2 + 1/n2≥ 2 for all n.
(b) Let xn:= 1/2nπ, un:= 1/(2nπ + π/2). Note that |g(xn)− g(un)| = 1 for all n.
4. Show that|f(x)−f(u)| ≤ [(|x|+|u|)/(1+x2)(1+u2)]|x−u| ≤ (1/2+1/2)|x−u| =
|x − u|. (Note that x → x/(1 + x2) attains a maximum of 1/2 at x = 1.) 5. Note that |(f(x) + g(x)) − (f(u) + g(u))| ≤ |f(x) − f(u)| + |g(x) − g(u)| < ε
provided that |x − u| < inf{δf(ε/2), δg(ε/2)}.
6. If M is a bound for both f and g on A, show that |f(x)g(x) − f(u)g(u)| ≤ M|f(x) − f(u)| + M|g(x) − g(u)| for all x, u ∈ A.
7. Since lim
x→0(sin x)/x = 1, there exists δ > 0 such that sin x≥ x/2 for 0 ≤ x < δ.
Let xn:= 2nπ and un:= 2nπ + 1/n, so that sin xn= 0 and sin un= sin(1/n).
If h(x) := x sin x, then|h(xn)− h(un)| = unsin(1/n)≥ (2nπ + 1/n)/2n > π > 0 for sufficiently large n.
8. Given ε > 0 there exists δf> 0 such that|y − v| < δf implies|f(y) − f(v)| < ε.
Now choose δg> 0 so that|x − u| < δg implies |g(x) − g(u)| < δf. 9. Note that|1/f(x) − 1/f(u)| ≤ (1/k2)|f(x) − f(u)|.
10. There exists δ > 0 such that if |x − u| < δ, x, u ∈ A, then |f(x) − f(u)| < 1. If A is bounded, it is contained in the finite union of intervals of length δ.
11. If|g(x) − g(0)| ≤ K|x − 0| for all x ∈ [0, 1], then√
x≤ Kx for x ∈ [0, 1]. But if xn:= 1/n2, then K must satisfy n≤ K for all n ∈ N, which is impossible.
12. Given ε > 0, choose 0 < δ1< 1 so that|f(x) − f(u)| < ε whenever |x − u| < δ1
and x, u∈ [0, a + 1]. Also choose 0 < δ2< 1 so that|f(x) − f(u)| < ε whenever
|x − u| < δ2 and x, u∈ [a, ∞). Now let δ := inf{δ1, δ2}. If |x − u| < δ, then since δ < 1, either x, u∈ [0, a + 1] or x, u, ∈ [a, ∞), so that |f(x) − f(u)| < ε.
13. Note that|f(x) − f(u)| ≤ |f(x) − gε(x)| + |gε(x)− gε(u)| + |gε(u)− f(u)|.
14. Since f is bounded on [0, p], it follows that it is bounded on R. Since f is continuous on J := [−1, p + 1], it is uniformly continuous on J. Now show that this implies that f is uniformly continuous on R.
15. Assume |f(x) − f(y)| ≤ Kf|x − y| and |g(x) − g(y)| ≤ Kg|x − y| for all x, y in A. (a) |(f(x) + g(x)) − (f(y) + g(y))| ≤ |f(x) − f(y)| + |g(x) − g(y)| ≤ (Kf+ Kg)|x + y|.
(b) If|f(x)| ≤ Bf and |g(x)| ≤ Bg for all x in A, then
|f(x)g(x) − f(y)g(y)| = |f(x)g(x) − f(x)g(y) + f(x)g(y) − f(y)g(y)|
≤ Bf|g(x) − g(y)| + Bg|f(x) − f(y)|
≤ (BfKg+ BgKf)|x − y|.
(c) Consider f (x) = x.
16. If|f(x) − f(y)| ≤ K|x − y| for all x, y in I, then has Lipschitz constant K on I.
Then for disjoint subintervals [xk,yk],n = 1,2,. . ., n, we have Σ|f(xk)−f(yk)| ≤ ΣK|xk− yk|, so that if ε > 0 is given and δ = ε/nK, then Σ|f(xk)− f(yk)| ≤ ε.
Thus f is absolutely continuous on I.
Section 5.5
In this section we introduce the notion of a “gauge” which will be used in the development of the generalized Riemann integral in Chapter 10. We will also use gauges to give alternate proofs of the main theorems in Section 5.3 and 5.4, Dini’s Theorem 8.2.6, and the Lebesgue Integrability Criterion in Appendix C.
Sample Assignment: Exercises 1, 2, 4, 6, 7, 9.
Partial Solutions:
1. (a) The δ-intervals are [0−14, 0 +14] = [−14,14], [12−14,12+14] = [14,34] and [34− 38,34+38] = [38,98].
(b) The third δ-interval is [103,109] which does not contain [12, 1].
2. (a) Yes. Since δ(t)≤ δ1(t), every δ-fine partition is δ1-fine.
(b) Yes. The third δ1-interval is [203,2120] which contains [12, 1].
3. No. The first δ2-interval is [−101,101] and does not contain [0,14].
4. (b) If t∈ (12, 1) then [t− δ(t), t + δ(t)] = [−12+32t,12+12t]⊂ (14, 1).
5. Routine verification.
6. We could have two subintervals having c as a tag with one of them not contained in the δ-interval around c. Consider constant gauges δ:= 1 on [0, 1] and δ:=12 on [1, 2], so that δ(1) =12. If ˙P consists of the single pair ([0, 1], 1), it is δ-fine. However, ˙P is not δ-fine.
7. Clearly δ∗(t) > 0 so that δ∗ is a gauge on [a, b]. If
P := {([a, x˙ 1], t1), . . . ([xk−1, c], tk), ([c, xk+1], tk+1), . . . , ([xn, b], tn)}
40 Bartle and Sherbert
is δ∗-fine, then it is clear that P˙:={([a, x1], t1), . . . , ([xk−1, c], tk)} is a δ-fine partition of [a, c] and P˙:={([c, xk+1], tk+1), . . . , ([xn, b], tn)} is a δ-fine partition of [c, b]. Evidently ˙P = ˙P∪ ˙P.
8. (a) If [a, b] a δ-fine partition ˙P and [c, b] has a δ-fine partition ˙P, then P˙∪ ˙P is a δ-fine partition of [a, b], contrary to hypothesis.
(b) Let I1 be [a, c] if it does not have a δ-fine partition; otherwise, let I1 be [c, b], so the length of I1 is (b− a)/2. Now bisect I1 and let I2, which has length (b− a)/22, be an interval that has no δ-fine partition. Continue this process by Induction.
(c) By the Nested Intervals Theorem there exists a common point ξ. By the Archimedean Property there exists p∈ N such that (b − a)/2p< δ(ξ). Since ξ∈ Ipand the length of Ipis (b− a)/2p, it follows that Ip⊂ [ξ − δ(ξ), ξ + δ(ξ)].
9. The hypothesis that f is locally bounded presents us with a gauge δ. If {([xi−1, xi], ti)}ni=1 is a δ-fine partition of [a, b] and Mi is a bound for |f| on [xi−1, xi], let M := sup{Mi : i = 1, . . . , n}.
10. The hypothesis that f is locally increasing presents us with a gauge δ. If {([xi−1, xi], ti)}ni=1 is a δ-fine partition of [a, b], then f is increasing on each interval [xi−1, xi]. By Induction it follows that f (xi)≤ f(xj) for i < j. If x < y belong to [a, b], then x∈ [xi−1, xi] and y∈ [xj−1, xj] where i≤ j. If i = j, the fact that f is increasing on [xi−1, xi] implies that f (x)≤ f(y). If i < j, then f (x)≤ f(xi)≤ f(xj−1)≤ f(y).
Section 5.6
The collection of monotone functions is a special, but very useful class of functions.
This is particularly the case since most functions that arise in elementary analysis are either monotone, or their domains can be written as a union of intervals on which their restrictions are monotone. Theorem 5.6.4 shows that a monotone function is automatically continuous except (at most) at a countable set of points.
It will also be seen in Theorem 5.6.5 that continuous strictly monotone functions have continuous strictly monotone inverse functions.
Sample Assignment: Exercises 1, 2, 4, 5, 7, 10, 12.
Partial Solutions:
1. If x∈ [a, b], then f(a) ≤ f(x).
2. If x1≤ x2, then f (x1)≤ f(x2) and g(x1)≤ g(x2), whence f (x1) + g(x1)≤ f (x2) + g(x2).
3. Note that (f g)(0) = 0 > (f g)(1/2) =−1/4.
4. If 0≤ f(x1)≤ f(x2) and 0≤ g(x1)≤ g(x2), then f (x1)g(x1)≤ f(x2)g(x1)≤ f (x2)g(x2).
5. If L := inf{f(x) : x ∈ (a, b]} and ε > 0, then there exists xε∈ (a, b] with L ≤ f (xε) < L + ε. Since f is increasing, then L≤ f(x) < L + ε for x ∈ (a, xε];
hence lim
x→a+ f exists and equals L.
Conversely, if K := lim
x→a+f , then given ε > 0, there exists δ > 0 such that if x∈ (a, a + δ), then K − ε < f(x) < K + ε. It follows from this that K− ε ≤ L < K + ε; since ε > 0 is arbitrary, we have K = L.
6. If f is continuous at c, then lim(f (xn)) = f (c), since c = lim(xn). Conversely, since 0≤ jf(c)≤ f(x2n)− f(x2n+1), it follows that jf(c) = 0, so f is continu-ous at c.
7. It follows from Exercises 2.4.4, 2.4.6 and the Principle of the Iterated Infima, (analogous to the result in Exercise 2.4.12), that
jf(c) = inf{f(y) : y ∈ I, c < y} − sup{f(x) : x ∈ I, x < c}
= inf{f(y) : y ∈ I, c < y} + inf{−f(x) : x ∈ I, x < c}
= inf{f(y) − f(x) : x, y ∈ I, x < c < y}
8. Let x1∈ I be such that y = f(x1) and x2∈ I be such that y = g(x2). If x2≤ x1, then y = g(y2) < f (x2)≤ f(x1) = y, a contradiction.
9. If x∈ I is rational, then f(x) = x is also rational so f(f(x)) = f(x) = x; if y∈ I is irrational, then f(y) = 1 − y is irrational so f(f(y)) = f(1 − y) = 1 − (1− y) = y. Suppose that x1 = x2, xj∈ I; if x1∈ Q and x2∈ Q, then f(x/ 1)∈ Q and f (x2) /∈ Q, which implies that f(x1) = f(x2). The other cases are similar.
Since|f(x) − 1/2| = |x − 1/2|, then f is continuous at 1/2. If x = 1/2, x ∈ Q, take a sequence (yn) of irrationals converging to x, so that f (yn) = 1− yn→ 1 − x = x. Similarly for the case x = 1/2, x /∈ Q.
10. If f has an absolute maximum at c∈ (a, b), and if f is injective, we have f (a) < f (c) and f (b) < f (c). Either f (a)≤ f(b) or f(b) < f(a). In the first case, either f (a) = f (b) or f (a) < f (b) < f (c), whence there exists b∈ (a, c) such that f (b) = f (b). Either possibility contradicts the assumption that f is injective. The case f (b) < f (a) is similar.
11. Note that f−1 is continuous at every point of its domain [0, 1]∪ (2, 3]. The function f is not continuous at x = 1.
12. Let a∈ (0, 1) be arbitrary. If f(a) < f(0), then there exists a∈ (a, 1) with f (a) = f (0), a contradiction. Also f (a) = f (0) is excluded by hypothesis.
Therefore we must have f (0) < f (a), and a similar argument yields f (a) < f (1).
If b∈ (a, 1) is given, then f(b) < f(a) implies that there exists a∈ (b, 1) with f (a) = f (a), a contradiction. Since f (b) = f (a) is excluded, we must have f (b) > f (a).
13. Assume that h is continuous on [0, 1] and let c1< c2 be the two points in [0, 1]
where h attains its supremum. If 0 < c1, choose a1, a2 such that 0 < a1< c1<
a2< c2. Let k satisfy sup{h(a1), h(a2)} < k < h(c1) = h(c2); then there exist
42 Bartle and Sherbert
three numbers bj such that a1< b1< c1< b2< a2< b3< c2 where k = h(bj), a contradiction. Now consider the points where h attains its infimum.
14. Let x > 0 and consider the case m, p, n, q∈ N. Let y := x1/n and z := x1/q so that yn= x = zq, whence (by Exercise 2.1.26) ynp= xp= zqp. Since np = mq, we have (ym)q= ymq= zpq= (zp)q, from which it follows that ym= zp, or (x1/n)m= (x1/q)p, or xm/n= xp/q. Now consider the case where m, p∈ Z.
15. Let x > 0 and consider the case where r = m/n and s = p/q, where m, n, p, q∈ N. Since r = mq/nq and s = pn/qn, it follows from the preceding exercise that xr= (x1/nq)mq and xs= (x1/nq)pn so that (by Exercise 2.1.26) xrxs= (x1/nq)mq + pn= x(mq+pn)/nq= xr+s. Similarly, xr= (x1/n)m> 0 and if y > 0, then (by 5.6.7) ys= (yp)1/q so that (xr)s= (((x1/n)m)p)1/q. This implies that ((xr)s)q= (x1/n)mp= (xmp)1/n so that ((xr)s)qn= xmp, whence (xr)s= xmp/qn= xrs. Now consider the case where m, p∈ Z.
DIFFERENTIATION
The basic properties and applications of the derivative are given in the first two sections of this chapter. Section 6.1 is a survey of the techniques of differentiation from a rigorous viewpoint. Since the students will be familiar with most of the results (though not the proofs), the section can be covered reasonably quickly.
Section 6.2 contains material that is new to students, since in introductory calculus courses the Mean Value Theorem is not usually given the emphasis it deserves.
Sections 6.3 and 6.4 are optional and can be discussed in either order and to whatever depth that time permits.
Section 6.1
This section contains the calculational rules of differentiation that students learn and use in introductory calculus courses. However, the emphasis here is on the rigorous establishment of these results rather than on the development of calcu-lational skills.
The topic that students will find troublesome is the differentiation of com-posite and inverse functions. We feel that the use of Carath´eodory’s Theorem 6.1.5 is a considerable simplification of the proofs of these results.
Sample Assignment: Exercises 1(a,b), 2, 4, 5, 9, 11, 13, 15.
Partial Solutions:
44 Bartle and Sherbert 4. Note that|f(x)/x| ≤ |x| for x ∈ R.
5. (a) f(x) = (1− x2)/(1 + x2)2, (b) g(x) = (x− 1)/√
5− 2x + x2,
(c) h(x) = m(sin xk)m−1(cos xk)(kxk−1), (d) k(x) = 2x sec2(x2).
6. The function f is continuous for n≥ 2 and is differentiable for n ≥ 3.
7. By definition g(c) = lim
h→0|f(c + h)|/h, if this limit exists. If 0 = |f(c)|=
hlim→0|f(c + h)/h|, it follows that g(c) = 0. If f(c) = L = 0, then we have lim(f (c± 1/n)/(±1/n)) = L, while lim(|f(c ± 1/n)|/(±1/n))] = ±L, so that
|f|(c) does not exist.
8. (a) f(x) = 2 for x > 0; f(x) = 0 for−1 < x < 0; and f(x) =−2 for x < −1, (b) g(x) = 3 if x > 0; g(x) = 1 if x < 0; g(0) does not exist,
(c) h(x) = 2|x| for all x ∈ R,
(d) k(x) = (−1)ncos x for nπ < x < (n + 1)π, n∈ Z; k(nπ) does not exist, (e) p(0) = 0; if x = 0, then p(x) does not exist.
9. If f is an even function, then f(−x) = lim
h→0[f (−x + h) − f(−x)]/h =
−limh→0[f (x− h) − f(x)]/(−h) = −f(x).
10. If x = 0, then g(x) = 2x sin(1/x2)− (2/x) cos(1/x2). Moreover, g(0) =
hlim→0h sin(1/h2) = 0. If xn:= 1/√
2nπ, then xn→ 0 and |g(xn)| = 2√ 2nπ, so g is unbounded in every neighborhood of 0.
11. (a) f(x) = 2/(2x + 3), (b) g(x) = 6(L(x2))2/x, (c) h(x) = 1/x, (d) k(x) = 1/(xL(x)).
12. r > 1.
13. Many examples are possible. For example, let f (x) := x for x rational and f (x) := 0 for x irrational.
14. 1/h(0) = 1/2, 1/h(1) = 1/5 and 1/h(−1) = 1/5.
15. D[Arccos y] = 1/D[cos x] =−1/ sin x = −1/ 1− y2. 16. D[Arctan y] = 1/D[tan x] = 1/ sec2x = 1/(1 + y2).
17. Given ε > 0, let δ(ε) > 0 be such that if 0 <|w − c| < δ(ε), w ∈ I, then
|f(w) − f(c) − (w − c)f(c)| < ε|w − c|. Now take w = u and w = v as described and subtract and add the term f (c)− f(c)c and use the Triangle Inequality to get
|f(v) − f(u) − f(c)(v− u)| ≤ |f(v) − f(c) − f(c)(v− c)|
+|f(c) − f(u) − f(c)(c− u)| ≤ ε|v − c| + ε|c − u|.
Since v− c ≥ 0 and c − u ≥ 0, then |v − c| = v − c and |c − u| = c − u, so that the final term equals ε(v− c + c − u) = ε(v − u).
Section 6.2
The Mean Value Theorem is stated for a function f on an interval [a, b]. However, many of its applications use intervals of the form [a, x] or [x1, x2] where x or x1, x2 are points in [a, b]. The shift from a “fixed interval” to what seems to be a
“variable interval” can cause confusion for some students. A word of explanation when this first occurs will help to alleviate this confusion.
WARNING: Exercises 16 and 18 are rather difficult.
Sample Assignment: Exercises 2(a, b), 3(a, b), 6, 7, 9, 10, 12, 13, 17.
Partial Solutions:
1. (a) Increasing on [3/2,∞), decreasing on (−∞, 3/2], (b) Increasing on (−∞, 3/8], decreasing on [3/8, ∞), (c) Increasing on (−∞, −1] and [1, ∞),
(d) Increasing on [0,∞).
2. (a) f(x) = 1−1/x2. Relative minimum at x = 1; relative maximum at x =−1, (b) g(x) = (1 + x)(1− x)/(1 + x2)2. Relative minimum at x =−1; relative maximum at x = 1,
(c) h(x) = 1/2√
x− 1/√
x + 2. Relative maximum at x = 2/3, (d) k(x) = 2(x3− 1)/x3. Relative minimum at x = 1.
3. (a) Relative minima at x =±1; relative maxima at x = 0, ±4, (b) Relative maximum at x = 1; relative minima at x = 0, 2, (c) Relative minima at x =−2, 3; relative maximum at x = 2,
(d) k(x) = 4(x− 6)/3(x − 8)2/3. Relative minimum at x = 6; relative maxima at x = 0, 9.
4. x = (1/n)(a1+· · · + an).
5. Show that f(x) < 0 for x > 1. Then f is strictly decreasing on [1,∞) so that f (a/b) < f (1) for a > b > 0.
6. If x < y, there exists c in (x, y) such that | sin x − sin y| = | cos c||y − x|.
7. There exists c with 1 < c < x such that ln x = (x− 1)/c. Now use the inequal-ity 1/x < 1/c < 1.
8. If h > 0 and a + h < b, there exists ch∈ (a, a + h) such that f(a + h) − f(a) = hf(ch). Since ch→ a as h → 0 + , it follows that f(a) = lim
h→0+[f (a + h)− f (a)]/h = lim
h→0+f(ch) = A. Now consider h < 0.
9. f (x) = x4(2 + sin(1/x)) > 0 for all x = 0, so f has an absolute minimum at x = 0. We have f(x) = 8x3+ 4x3sin(1/x)− x2cos(1/x) for x = 0. Now verify that f(1/2nπ) < 0 for n≥ 2 and f(2/(4n + 1)π) > 0 for n≥ 1.
46 Bartle and Sherbert 10. g(0) = lim
x→0(1 + 2x sin(1/x)) = 1 + 0 = 1, and if x = 0, then g(x) = 1 + 4x sin(1/x)− 2 cos(1/x). Now show that g(1/2nπ) < 0 and that we have g(2/(4n + 1)π) > 0 for n∈ N.
11. For example, f (x) := √ x.
12. Apply Darboux’s Theorem 6.2.12. If g(x) := a for x < 0, g(x) := x + b for x≥ 0, where a, b are any constants, then g(x) = h(x) for x = 0.
13. If x1< x2, then there exists c∈ (x1, x2) such that f (x2)− f(x1) = (x2− x1)f(c) > 0.
14. Apply Darboux’s Theorem 6.2.12.
15. Suppose that |f(x)| ≤ K for x ∈ I. For x, y ∈ I, apply the Mean Value Theorem to get |f(x) − f(y)| = |(x − y)f(c)| ≤ K|x − y|.
16. (a) Given ε > 0 there exists nε∈ N such that if x ≥ nε, then |f(x)− b| < ε.
Hence if x≥ nε and h > 0, there exists yx∈ (x, x + h) such that f (x + h)− f(x)
h − b
= |f(yx)− b| < ε.
Since ε > 0 is arbitrary, then lim
x→∞(f (x + h)− f(x))/h = b.
(b) Assume that b = 0 and let ε < |b|/2. Let nε be as in part (a). Since lim
x→∞f exists, we may also assume that if x, y≥ nε, then |f(x) − f(y)| < ε. Hence there exists xε in (nε, nε+ 1) such that ε >|f(nε+ 1)− f(nε)| = |f(xε)| ≥
|b|/2. Since ε > 0 is arbitrary, the hypothesis that b = 0 is contradicted.
(c) If x≥ nε, then there exists yε∈ (nε, x) such that f (x)− f(nε) = (x− nε)f(yε), so that we have
f (x)/x− b = f(yε)− b + f(nε)/x− nεf(yε)/x.
Since yε> nε, we have |f(yε)− b| < ε. Moreover |f(nε)/x| < ε if x is suffi-ciently large; since f is bounded on [nε,∞), then |nεf(yε)/x| < ε if x is sufficiently large. Therefore, lim
x→∞f (x)/x = b.
17. Apply the Mean Value Theorem to the function g− f on [0, x].
18. Given ε > 0, let δ = δ(ε) be as in Definition 6.1.1, and let x < c < y be such that 0 <|x − y| < δ. Since f(x) − f(y) = f(x) − f(c) + f(c) − f(y), a simple calculation shows that
f (x)− f(y)
x− y = x− c
x− y ·f (x)− f(c)
x− c + c− y
x− y ·f (y)− f(c) y− c .
Since both (x− c)/(x − y) and (c − y)/(x − y) are positive and have sum 1,
Note that if one (but not both) of x and y equal c, the conclusion still holds.
19. Let x, y∈ I, x = y; then If f is uniformly differentiable on I, given ε > 0 there exists δ > 0 such that if 0 <|x − y| < δ, x, y ∈ I, then both terms on the right side are less than ε.
Hence we have |f(x)− f(y)| < 2ε for |x − y| < δ, x, y ∈ I, whence we con-clude that f is (uniformly) continuous on I.
20. (a,b) Apply the Mean Value Theorem.
(c) Apply Darboux’s Theorem to the results of (a) and (b).
Section 6.3
The proofs of the various cases of L’Hospital’s Rules range from fairly trivial to rather complicated. The only really difficult argument in this section is the proof of Theorem 6.3.5, which deals with the case∞/∞. This requires a more subtle analysis than the other cases.
This section may be regarded as optional. Students are already familiar with the mechanics of L’Hospital’s Rules.
Sample Assignment: Exercises 1, 2, 4, 6, 7(a,b), 8(a,b), 9(a,b), 13, 14.
Partial Solutions:
48 Bartle and Sherbert
Add h(y)/ex to all sides and rearrange terms to get
(L− ε/2) + h(y)− ey(L− ε/2) values of x, then the above argument can be modified slightly to prove the
following version of L’Hospital’s Rule: If h and g are differentiable func-tions on (0,∞) that satisfy lim
x→∞h(x)/g(x) = L and lim
The applications of Taylor’s Theorem are similar in spirit to those of the Mean Value Theorem, but the technical details can be more complicated since higher order derivatives are involved. Instead of estimating f, the use of Taylor’s Theorem usually requires the estimation of the remainder Rn.
The applications that are presented here are independent of one another and
The applications that are presented here are independent of one another and